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a, x4 + 2x3 +x2 = x4 +x3 +x3 +x2 =(x4+x3 )+(x3 +x2 ) =x3(x +1 ) + x2 (x+1 ) =(x+1)(x3+x2)
a) x4 + 2x3 + x2
= x2(x2 + 2x + 1)
= x2(x + 1)2
= [x(x + 1)]2
= (x2 + x)2
b) 5x3 - 10xy + 5y2 - 20z2
= 5(x3 - 2xy + y2 - 4z2)
c) x2y - xy2 + x3 - y3
= xy(x - y) + (x - y)(x2 + xy + y2)
= (x - y)(x2 + 2xy + y2)
= (x - y)(x + y)2
d) x2 - xy + 4x - 2y + 4
= (x2 + 4x + 4) - (xy + 2y)
= (x + 2)2 - y(x + 2)
= (x + 2)(x + 2 - y)
d) x2 - x - 6
= x2 - 3x + 2x - 6
= x(x - 3) + 2(x - 3)
= (x + 2)(x - 3)
f) 3x2 - 5x - 8
= 3x2 + 3x - 8x - 8
= 3x(x + 1) - 8(x + 1)
= (3x - 8)(x + 1)
g) x3 + 3x2 + 6x + 4
= (x3 + 3x2 + 3x + 1) + (3x + 3)
= (x + 1)3 + 3(x + 1)
= (x + 1)[(x + 1)2 + 3]
h) 3x3 - 5x2 - 6x + 8
= 3x3 - 3x2 - 2x2 - 6x + 8
= 3x3 - 3x2 - 2x2 + 2x - 8x + 8
= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)
= (3x2 - 2x - 8)(x - 1)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) \(x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
b) \(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
c) \(2x^3-x^2-8x+4\)
\(=x^2\left(2x-1\right)-4\left(2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)
d) \(x\left(x-y\right)^2+y\left(x-y\right)^2-xy+x^2\)
\(=\left(x+y\right)\left(x-y\right)^2+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2+x\right)\)
e) \(2x^2-5x+2\)
\(=\left(2x^2-x\right)-\left(4x-2\right)\)
\(=x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
b) x3y3 + x2y2+ 4 = x3y3- 4xy + (xy)2- 2xy.2 + 22 = xy [ (xy)^2 - 2^2 ] + ( xy - 2)^2
= xy(xy-2)(xy+2)+ (xy-2)^2
= (xy-2) [ xy(xy+2) + ( xy-2) ]
= (xy-2) [ (xy)2 + 2xy + xy - 3 ]
= ( xy - 3) [ (xy)2 + 3xy - 3]
3) (chưa bik làm)
4) x4 +x3 + 6x2 +5x +5
= x4 +x3 + x2 + 5x2 + 5x +5
= x2( x2+x+ 1 ) + 5( x2+x+ 1 )
= ( x2+ 5 ) ( x2+x+ 1 )
5) x4 - 2x3 - 12x2 +12x + 36
= x4 - 2x3 - 6x2 - 6x2 + 12x + 36=
x2 ( x2 - 2x - 6) - 6 ( x2 - 2x - 6)
= (x^2 - 6) ( x2 - 2x - 6) 6) x8y8 + x4y4 + 1 = \(\left[\left(xy\right)^4\right]^2+2x^4y^4+1-x^4y^4\)=\(\left[\left(xy\right)^4+1\right]^2-\left[\left(xy\right)^2\right]^2\)
= \(\left(x^4y^4+1-x^2y^2\right)\left(x^4y^4+1+x^2y^2\right)\)
( mik ko bik đúng hay sai đâu nha) mik thấy nó thành nhân tử thì mik tách thôi
Câu a trước đi ạ ^^
a) 7x - 6x2 - 2
= - 6x2 + 7x - 2
= (- 6x2 + 3x) + (4x - 2)
= 3x (- 2x + 1) + 2 (2x-1)
= - 3x ( 2x -1) + 2 (2x - 1)
= ( 2x -1 ) ( - 3x +2 )
Bài làm
a) 4x2 - 6x
= 2x( 2x - 3 )
b) 9x4y3 + 3x2y4
= 3x2y3( 3x2 + y )
c) x3 - 2x2 + 5x
= x( x2 - 2x + 5 )
d) 3x( x - 1 ) + 5( x - 1 )
= ( x - 1 )( 3x + 5 )
e) 2x2( x + 1 ) + 4( x + 1 )
= ( x + 1 )( 2x2 + 4 )
= ( x + 1 )2( x2 + 2 )
= 2( x + 1 )( x2 + 2 )
f) -3x - 6xy + 9xz
= -( 3x + 6xy - 9xz )
= -3x( 1 + 2y - 3z )
# Học tốt #
a) \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(x^2y-xy^2+x^3-y^3\)
\(=xy\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\)
\(=\left(x+2\right)^2-y\left(x+2\right)\)
\(=\left(x+2\right)\left(x-y+2\right)\)
e) \(x^2-x-6=\left(x+2\right)\left(x-3\right)\)
f) \(3x^2-5x-8\)
\(=\left(3x^2+3x\right)-\left(8x+8\right)\)
\(=3x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-8\right)\)